Integrate 2x + 1/(x^2 + 4) dx

pls integrate using partial fractions

I'm pretty sure you can't use partial fractions on this. You're actually gonna have to use a little bit of trig integration on this one.

∫(2x + 1) / (x² + 4) dx

Break this up into 2 separate integrals....

∫2xdx / (x² + 4) + ∫dx / (x² + 4)

For the first integral, use a u substitution....
u = x² + 4
du = 2x dx

∫du / u

Integrate like normal....

ln[u]

Substitute x² + 4 back in for u....

ln[x² + 4]
For the second integral, there's an identity that says that whenever you have the integral of dx/(x² + a), the answer will always be the (arctan(x /a)) / a. For our problem, we have the integral of ∫dx / (x² + 4), so the integral of that will be (1/2*arctan(x / 2).

Putting everything together....

Final Answer:
ln[x² + 4] + (1/2)*arctan(x / 2) + C

Question not clear.
Case 1
∫[2x + {1/(x^2+4)}]dx = x^2 + (1/2)tan^-1(x/2)

Case 2
∫[(2x +1)/(x^2+4)}]dx = ∫{[(2x)/(x^2+4)}]dx=I1say} + ∫[1/(x^2+4)}]dx
I1 + (1/2)tan^-1(x/2), and I1 is given by
I1 = ln(x^2+4), because d/dx{(x^2+4)} = 2x. So
∫[(2x +1)/(x^2+4)}]dx = ln(x^2+4) +(1/2)tan^-1(x/2)

Answer: 1/2ln(x) + c

Oh, partial fractions.

Eh. I'm lazy.